NYJC 2023 H2 Physics 9749 P3 Answer

NYJC 2023 9749/03/J2Prelim/23 [Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME Solution CLASS TUTOR’S NAME CENTRE NUMBER S INDEX NUMBER PHYSICS 9749/03 Paper 3 Longer Structured Questions 20 September 2023 2 hours Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class, Centre number and index number in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Section A Answer all questions. Section B Answer one question only. You are advised to spend one and a half hours on Section A and half an hour on Section B. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 / 8 2 / 8 3 / 8 4 / 8 5 / 8 6 / 9 7 / 11 Section B 8 / 20 9 / 20 Total / 80 This document consists of 24 printed pages.

2 NYJC 2023 9749/03/J2Prelim/23 Data speed of light in free space c = 3.00 × 108 m s−1 permeability of free space  = 4 × 10−7 H m−1 permittivity of free space  = 8.85 × 10−12 F m−1 (1 / (36)) × 10−9 F m−1 elementary charge e = 1.60 × 10−19 C the Planck constant h = 6.63 × 10−34 J s unified atomic mass constant u = 1.66 × 10−27 kg rest mass of electron me = 9.11 × 10−31 kg rest mass of proton mp = 1.67 × 10−27 kg molar gas constant R = 8.31 J K−1 mol−1 the Avogadro constant NA = 6.02 × 1023 mol−1 the Boltzmann constant k = 1.38 × 10−23 J K−1 gravitational constant G = 6.67 × 10−11 N m2 kg−2 acceleration of free fall g = 9.81 m s−2 Formulae uniformly accelerated motion 21 2s ut at=+ 22 2v u as=+ work done on / by a gas W p V= hydrostatic pressure p gh= gravitational potential /Gm r =− temperature / K / C 273.15TT =  + pressure of an ideal gas 21 3 Nmpc V=   mean translational kinetic energy of an ideal molecule 3 2E kT= displacement of particle in s.h.m. 0 sinx x t = velocity of particle in s.h.m. 0 cosv v t = 22 0xx=  − electric current =I Anvq resistors in series 12 . . .R R R= + + resistors in parallel 121/ 1/ 1/ . . .R R R= + + electric potential 04 QV r= alternating current/voltage 0 sinx x t = magnetic flux density due to a long straight wire  = 0 2 IB d magnetic flux density due to a flat circular coil = 0 2 NIB r magnetic flux density due to a long solenoid = 0B nI radioactive decay 0 exp( )x x t =− decay constant 1 2 ln2 t =

3 NYJC 2023 9749/03/J2Prelim/23 [Turn over Section A Answer all the questions in this section in the spaces provided. 1 (a) A student sets up the circuit shown in Fig . 1.1 to determine the resistivity of the metal of a resistance wire. He reads the current I from the ammeter and the potential difference V from the voltmeter. Fig. 1.1 (i) Determine the base units for resistance.   2 22 -2 2 2 -3 -2 kg m s m sA kg m s A P I R P FdR I I t = = = =    =       base units for resistance: ………………………. [2] (ii) The following readings were obtained for the experiment: Reading of voltmeter = 1.30 ± 0.01 V Reading of ammeter = 0.76 ± 0.01 A Length of wire = 75.4 ± 0.2 cm Diameter of wire = 0.54 ± 0.02 mm Calculate, with its associated uncertainty, the value of the resistivity of the metal of the resistance wire, expressing your results to an appropriate number of significant figures. 2 23 [ 2 2 C1]7 4 (1.30)(0.54 10 ) 4 4(0.76)(75.4 10 ) 5.19 10 m LR A VL dI Vd IL     − − − = = ==  =   [M1] [A1] [A1] 7 7 7 2 0.01 0.01 0.02 0.22 1.30 0.76 0.54 75.4 0.098 (5.19 10 ) 0.5 10 5.2 0.5 10 m V I d L V I d L     − − −     = + + + = + + +  =   = =    ρ = ………………………. ± ………………………. Ω m [4] A V resistance wire

4 NYJC 2023 9749/03/J2Prelim/23 (b) A second student repeated the experiment in (a) with the same length of the wire. In this new experiment, the supply voltage was varied and several pairs of corresponding readings on the voltmeter and ammeter were tabulated. A graph showing the variation of the current in the wire with potential difference across the wire was then plotted. Discuss how this procedure reduces the random error and systematic error that could have occurred. ...…...……………………………………………….………………………………………………… …...…………………………………………………………………………………………………… ………...………………………………………………………………………………………….. [2] [Total: 8] 2 An object of mass 1.5 kg is released from a stationary hot air balloon. Fig. 2.1 shows how the vertical displacement of the object varies with time. Fig. 2.1 (a) Calculate the change in gravitational potential energy ∆Ep of the object that occurred during the 16 s after it was released. ∆Ep = ……………….. J [1] 0 100 200 300 400 500 600 0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 ∆Ep = mgh = 1.5 (9.81)(-510) = (-) 7500 J [A1] vertical displacement / m time / s The graph of best fit will reduce the scatter caused by random errors in the experiment. [B1] If the line of best fit does not pass through the origin, there is a systematic error present in the experiment. [B1]

5 NYJC 2023 9749/03/J2Prelim/23 [Turn over (b) Using Fig. 2.1, determine the speed of the object at t = 16 s. speed = ……………….. m s-1 [2] (c) Calculate the change in kinetic energy ∆Ek of the object during the same period. ∆Ek = ………………….. J [1] (d) Explain why ∆Ep and ∆Ek are not equal to one another. …..........……………………………………………………………………………………………… ………...………………………………………………………………………………………….. [1] (e) The object strikes the ground 16 s after it was released and penetrates 0.85 m into the ground. Determine the average resistive force acting on the object as it penetrates the ground. average resistive force = ………………….. N [3] [Total: 8] ∆EK = ½mv2 - ½mu2 = ½ (1.5)(50)2 – 0 = 1880 J [A1] Work is done (by object) against/to overcome air resistance. [B1] Work done against resistive force = Loss of EK (and EP) F(0.85) = 1880 [M1] (+ 1.5×9.81×0.85 [B1]) F = 2210 N (or 2230 N) [A1] ( )( ) [M1] [B1] [A1] 250.01.5 9.81 1.5 2 0.85 2210 N (or 2230 N) f mg ma f −= − + =   = speed = gradient of tangent to the graph at t = 16 s [M1] = 50 m s-1 (acceptable range: 49 m s-1 to 52 m s-1) [A1]

6 NYJC 2023 9749/03/J2Prelim/23 mass spring fixed point air track fixed point 100 80 60 energy / mJ 40 20 –6.0 –4.0 –2.0 0 2.0 4.0 6.0 x / cm 3 Fig. 3.1 shows a 0.45 kg mass held on a horizontal air track by two identical springs that are initially unstretched. The mass is pulled 5.0 cm to the left and released. The mass oscillates horizontally on a cushion of air. Fig. 3.1 (a) Explain why the mass oscillates with simple harmonic motion when displaced horizontally. …...…………………………………………………………………………………………………… ...…...……………………………………………….………………………………………………… …...…………………………………………………………………………………………………… ………...………………………………………………………………………………………….. [2] The variation of kinetic energy with displacement x of the mass is shown in Fig. 3.2. Fig. 3.2 Resultant force on mass is proportional to displacement from equilibrium position / fixed point (as springs obey Hooke’s law) so acceleration of